\(\int \frac {a+b x+c x^2}{x^4 \sqrt {1-d x} \sqrt {1+d x}} \, dx\) [7]

Optimal result

Integrand size = 33, antiderivative size = 99 \[ \int \frac {a+b x+c x^2}{x^4 \sqrt {1-d x} \sqrt {1+d x}} \, dx=-\frac {a \sqrt {1-d^2 x^2}}{3 x^3}-\frac {b \sqrt {1-d^2 x^2}}{2 x^2}-\frac {\left (3 c+2 a d^2\right ) \sqrt {1-d^2 x^2}}{3 x}-\frac {1}{2} b d^2 \text {arctanh}\left (\sqrt {1-d^2 x^2}\right ) \] Output:

-1/3*a*(-d^2*x^2+1)^(1/2)/x^3-1/2*b*(-d^2*x^2+1)^(1/2)/x^2-1/3*(2*a*d^2+3* 
c)*(-d^2*x^2+1)^(1/2)/x-1/2*b*d^2*arctanh((-d^2*x^2+1)^(1/2))
 

Mathematica [A] (verified)

Time = 0.24 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.78 \[ \int \frac {a+b x+c x^2}{x^4 \sqrt {1-d x} \sqrt {1+d x}} \, dx=\frac {1}{6} \left (-\frac {\sqrt {1-d^2 x^2} \left (3 x (b+2 c x)+a \left (2+4 d^2 x^2\right )\right )}{x^3}-3 b d^2 \log (x)+3 b d^2 \log \left (-1+\sqrt {1-d^2 x^2}\right )\right ) \] Input:

Integrate[(a + b*x + c*x^2)/(x^4*Sqrt[1 - d*x]*Sqrt[1 + d*x]),x]
 

Output:

(-((Sqrt[1 - d^2*x^2]*(3*x*(b + 2*c*x) + a*(2 + 4*d^2*x^2)))/x^3) - 3*b*d^ 
2*Log[x] + 3*b*d^2*Log[-1 + Sqrt[1 - d^2*x^2]])/6
 

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.06, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {2112, 2338, 25, 539, 25, 534, 243, 73, 221}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {a+b x+c x^2}{x^4 \sqrt {1-d x} \sqrt {d x+1}} \, dx\)

\(\Big \downarrow \) 2112

\(\displaystyle \int \frac {a+b x+c x^2}{x^4 \sqrt {1-d^2 x^2}}dx\)

\(\Big \downarrow \) 2338

\(\displaystyle -\frac {1}{3} \int -\frac {3 b+\left (2 a d^2+3 c\right ) x}{x^3 \sqrt {1-d^2 x^2}}dx-\frac {a \sqrt {1-d^2 x^2}}{3 x^3}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {1}{3} \int \frac {3 b+\left (2 a d^2+3 c\right ) x}{x^3 \sqrt {1-d^2 x^2}}dx-\frac {a \sqrt {1-d^2 x^2}}{3 x^3}\)

\(\Big \downarrow \) 539

\(\displaystyle \frac {1}{3} \left (-\frac {1}{2} \int -\frac {3 b x d^2+2 \left (2 a d^2+3 c\right )}{x^2 \sqrt {1-d^2 x^2}}dx-\frac {3 b \sqrt {1-d^2 x^2}}{2 x^2}\right )-\frac {a \sqrt {1-d^2 x^2}}{3 x^3}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {1}{3} \left (\frac {1}{2} \int \frac {3 b x d^2+2 \left (2 a d^2+3 c\right )}{x^2 \sqrt {1-d^2 x^2}}dx-\frac {3 b \sqrt {1-d^2 x^2}}{2 x^2}\right )-\frac {a \sqrt {1-d^2 x^2}}{3 x^3}\)

\(\Big \downarrow \) 534

\(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (3 b d^2 \int \frac {1}{x \sqrt {1-d^2 x^2}}dx-\frac {2 \sqrt {1-d^2 x^2} \left (2 a d^2+3 c\right )}{x}\right )-\frac {3 b \sqrt {1-d^2 x^2}}{2 x^2}\right )-\frac {a \sqrt {1-d^2 x^2}}{3 x^3}\)

\(\Big \downarrow \) 243

\(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (\frac {3}{2} b d^2 \int \frac {1}{x^2 \sqrt {1-d^2 x^2}}dx^2-\frac {2 \sqrt {1-d^2 x^2} \left (2 a d^2+3 c\right )}{x}\right )-\frac {3 b \sqrt {1-d^2 x^2}}{2 x^2}\right )-\frac {a \sqrt {1-d^2 x^2}}{3 x^3}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (-3 b \int \frac {1}{\frac {1}{d^2}-\frac {x^4}{d^2}}d\sqrt {1-d^2 x^2}-\frac {2 \sqrt {1-d^2 x^2} \left (2 a d^2+3 c\right )}{x}\right )-\frac {3 b \sqrt {1-d^2 x^2}}{2 x^2}\right )-\frac {a \sqrt {1-d^2 x^2}}{3 x^3}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (-\frac {2 \sqrt {1-d^2 x^2} \left (2 a d^2+3 c\right )}{x}-3 b d^2 \text {arctanh}\left (\sqrt {1-d^2 x^2}\right )\right )-\frac {3 b \sqrt {1-d^2 x^2}}{2 x^2}\right )-\frac {a \sqrt {1-d^2 x^2}}{3 x^3}\)

Input:

Int[(a + b*x + c*x^2)/(x^4*Sqrt[1 - d*x]*Sqrt[1 + d*x]),x]
 

Output:

-1/3*(a*Sqrt[1 - d^2*x^2])/x^3 + ((-3*b*Sqrt[1 - d^2*x^2])/(2*x^2) + ((-2* 
(3*c + 2*a*d^2)*Sqrt[1 - d^2*x^2])/x - 3*b*d^2*ArcTanh[Sqrt[1 - d^2*x^2]]) 
/2)/3
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 221

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x 
/Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
 

rule 243

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2   Subst[In 
t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I 
ntegerQ[(m - 1)/2]
 

rule 534

Int[(x_)^(m_)*((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> 
Simp[(-c)*x^(m + 1)*((a + b*x^2)^(p + 1)/(2*a*(p + 1))), x] + Simp[d   Int[ 
x^(m + 1)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, m, p}, x] && ILtQ[m, 
0] && GtQ[p, -1] && EqQ[m + 2*p + 3, 0]
 

rule 539

Int[(x_)^(m_)*((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> 
Simp[c*x^(m + 1)*((a + b*x^2)^(p + 1)/(a*(m + 1))), x] + Simp[1/(a*(m + 1)) 
   Int[x^(m + 1)*(a + b*x^2)^p*(a*d*(m + 1) - b*c*(m + 2*p + 3)*x), x], x] 
/; FreeQ[{a, b, c, d, p}, x] && ILtQ[m, -1] && GtQ[p, -1] && IntegerQ[2*p]
 

rule 2112

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f 
_.)*(x_))^(p_.), x_Symbol] :> Int[Px*(a*c + b*d*x^2)^m*(e + f*x)^p, x] /; F 
reeQ[{a, b, c, d, e, f, m, n, p}, x] && PolyQ[Px, x] && EqQ[b*c + a*d, 0] & 
& EqQ[m, n] && (IntegerQ[m] || (GtQ[a, 0] && GtQ[c, 0]))
 

rule 2338

Int[(Pq_)*((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{ 
Q = PolynomialQuotient[Pq, c*x, x], R = PolynomialRemainder[Pq, c*x, x]}, S 
imp[R*(c*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] + Simp[1/(a*c*( 
m + 1))   Int[(c*x)^(m + 1)*(a + b*x^2)^p*ExpandToSum[a*c*(m + 1)*Q - b*R*( 
m + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x] && Lt 
Q[m, -1] && (IntegerQ[2*p] || NeQ[Expon[Pq, x], 1])
 

Maple [A] (verified)

Time = 0.70 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.27

method	result	size
risch	\(\frac {\sqrt {x d +1}\, \left (x d -1\right ) \left (4 a \,d^{2} x^{2}+6 c \,x^{2}+3 b x +2 a \right ) \sqrt {\left (-x d +1\right ) \left (x d +1\right )}}{6 x^{3} \sqrt {-\left (x d +1\right ) \left (x d -1\right )}\, \sqrt {-x d +1}}-\frac {b \,d^{2} \operatorname {arctanh}\left (\frac {1}{\sqrt {-d^{2} x^{2}+1}}\right ) \sqrt {\left (-x d +1\right ) \left (x d +1\right )}}{2 \sqrt {-x d +1}\, \sqrt {x d +1}}\)	\(126\)
default	\(-\frac {\sqrt {-x d +1}\, \sqrt {x d +1}\, \operatorname {csgn}\left (d \right )^{2} \left (3 \,\operatorname {arctanh}\left (\frac {1}{\sqrt {-d^{2} x^{2}+1}}\right ) b \,d^{2} x^{3}+4 \sqrt {-d^{2} x^{2}+1}\, a \,d^{2} x^{2}+6 \sqrt {-d^{2} x^{2}+1}\, c \,x^{2}+3 \sqrt {-d^{2} x^{2}+1}\, b x +2 \sqrt {-d^{2} x^{2}+1}\, a \right )}{6 \sqrt {-d^{2} x^{2}+1}\, x^{3}}\)	\(130\)

Input:

int((c*x^2+b*x+a)/x^4/(-d*x+1)^(1/2)/(d*x+1)^(1/2),x,method=_RETURNVERBOSE 
)
 

Output:

1/6*(d*x+1)^(1/2)*(d*x-1)*(4*a*d^2*x^2+6*c*x^2+3*b*x+2*a)/x^3/(-(d*x+1)*(d 
*x-1))^(1/2)*((-d*x+1)*(d*x+1))^(1/2)/(-d*x+1)^(1/2)-1/2*b*d^2*arctanh(1/( 
-d^2*x^2+1)^(1/2))*((-d*x+1)*(d*x+1))^(1/2)/(-d*x+1)^(1/2)/(d*x+1)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.79 \[ \int \frac {a+b x+c x^2}{x^4 \sqrt {1-d x} \sqrt {1+d x}} \, dx=\frac {3 \, b d^{2} x^{3} \log \left (\frac {\sqrt {d x + 1} \sqrt {-d x + 1} - 1}{x}\right ) - {\left (2 \, {\left (2 \, a d^{2} + 3 \, c\right )} x^{2} + 3 \, b x + 2 \, a\right )} \sqrt {d x + 1} \sqrt {-d x + 1}}{6 \, x^{3}} \] Input:

integrate((c*x^2+b*x+a)/x^4/(-d*x+1)^(1/2)/(d*x+1)^(1/2),x, algorithm="fri 
cas")
 

Output:

1/6*(3*b*d^2*x^3*log((sqrt(d*x + 1)*sqrt(-d*x + 1) - 1)/x) - (2*(2*a*d^2 + 
 3*c)*x^2 + 3*b*x + 2*a)*sqrt(d*x + 1)*sqrt(-d*x + 1))/x^3
 

Sympy [F(-1)]

Timed out. \[ \int \frac {a+b x+c x^2}{x^4 \sqrt {1-d x} \sqrt {1+d x}} \, dx=\text {Timed out} \] Input:

integrate((c*x**2+b*x+a)/x**4/(-d*x+1)**(1/2)/(d*x+1)**(1/2),x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.09 \[ \int \frac {a+b x+c x^2}{x^4 \sqrt {1-d x} \sqrt {1+d x}} \, dx=-\frac {1}{2} \, b d^{2} \log \left (\frac {2 \, \sqrt {-d^{2} x^{2} + 1}}{{\left | x \right |}} + \frac {2}{{\left | x \right |}}\right ) - \frac {2 \, \sqrt {-d^{2} x^{2} + 1} a d^{2}}{3 \, x} - \frac {\sqrt {-d^{2} x^{2} + 1} c}{x} - \frac {\sqrt {-d^{2} x^{2} + 1} b}{2 \, x^{2}} - \frac {\sqrt {-d^{2} x^{2} + 1} a}{3 \, x^{3}} \] Input:

integrate((c*x^2+b*x+a)/x^4/(-d*x+1)^(1/2)/(d*x+1)^(1/2),x, algorithm="max 
ima")
 

Output:

-1/2*b*d^2*log(2*sqrt(-d^2*x^2 + 1)/abs(x) + 2/abs(x)) - 2/3*sqrt(-d^2*x^2 
 + 1)*a*d^2/x - sqrt(-d^2*x^2 + 1)*c/x - 1/2*sqrt(-d^2*x^2 + 1)*b/x^2 - 1/ 
3*sqrt(-d^2*x^2 + 1)*a/x^3
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 619 vs. \(2 (83) = 166\).

Time = 0.26 (sec) , antiderivative size = 619, normalized size of antiderivative = 6.25 \[ \int \frac {a+b x+c x^2}{x^4 \sqrt {1-d x} \sqrt {1+d x}} \, dx=-\frac {3 \, b d^{3} \log \left ({\left | -\frac {\sqrt {2} - \sqrt {-d x + 1}}{\sqrt {d x + 1}} + \frac {\sqrt {d x + 1}}{\sqrt {2} - \sqrt {-d x + 1}} + 2 \right |}\right ) - 3 \, b d^{3} \log \left ({\left | -\frac {\sqrt {2} - \sqrt {-d x + 1}}{\sqrt {d x + 1}} + \frac {\sqrt {d x + 1}}{\sqrt {2} - \sqrt {-d x + 1}} - 2 \right |}\right ) + \frac {4 \, {\left (6 \, a d^{4} {\left (\frac {\sqrt {2} - \sqrt {-d x + 1}}{\sqrt {d x + 1}} - \frac {\sqrt {d x + 1}}{\sqrt {2} - \sqrt {-d x + 1}}\right )}^{5} - 3 \, b d^{3} {\left (\frac {\sqrt {2} - \sqrt {-d x + 1}}{\sqrt {d x + 1}} - \frac {\sqrt {d x + 1}}{\sqrt {2} - \sqrt {-d x + 1}}\right )}^{5} - 16 \, a d^{4} {\left (\frac {\sqrt {2} - \sqrt {-d x + 1}}{\sqrt {d x + 1}} - \frac {\sqrt {d x + 1}}{\sqrt {2} - \sqrt {-d x + 1}}\right )}^{3} + 6 \, c d^{2} {\left (\frac {\sqrt {2} - \sqrt {-d x + 1}}{\sqrt {d x + 1}} - \frac {\sqrt {d x + 1}}{\sqrt {2} - \sqrt {-d x + 1}}\right )}^{5} + 96 \, a d^{4} {\left (\frac {\sqrt {2} - \sqrt {-d x + 1}}{\sqrt {d x + 1}} - \frac {\sqrt {d x + 1}}{\sqrt {2} - \sqrt {-d x + 1}}\right )} - 48 \, c d^{2} {\left (\frac {\sqrt {2} - \sqrt {-d x + 1}}{\sqrt {d x + 1}} - \frac {\sqrt {d x + 1}}{\sqrt {2} - \sqrt {-d x + 1}}\right )}^{3} + 48 \, b d^{3} {\left (\frac {\sqrt {2} - \sqrt {-d x + 1}}{\sqrt {d x + 1}} - \frac {\sqrt {d x + 1}}{\sqrt {2} - \sqrt {-d x + 1}}\right )} + 96 \, c d^{2} {\left (\frac {\sqrt {2} - \sqrt {-d x + 1}}{\sqrt {d x + 1}} - \frac {\sqrt {d x + 1}}{\sqrt {2} - \sqrt {-d x + 1}}\right )}\right )}}{{\left ({\left (\frac {\sqrt {2} - \sqrt {-d x + 1}}{\sqrt {d x + 1}} - \frac {\sqrt {d x + 1}}{\sqrt {2} - \sqrt {-d x + 1}}\right )}^{2} - 4\right )}^{3}}}{6 \, d} \] Input:

integrate((c*x^2+b*x+a)/x^4/(-d*x+1)^(1/2)/(d*x+1)^(1/2),x, algorithm="gia 
c")
 

Output:

-1/6*(3*b*d^3*log(abs(-(sqrt(2) - sqrt(-d*x + 1))/sqrt(d*x + 1) + sqrt(d*x 
 + 1)/(sqrt(2) - sqrt(-d*x + 1)) + 2)) - 3*b*d^3*log(abs(-(sqrt(2) - sqrt( 
-d*x + 1))/sqrt(d*x + 1) + sqrt(d*x + 1)/(sqrt(2) - sqrt(-d*x + 1)) - 2)) 
+ 4*(6*a*d^4*((sqrt(2) - sqrt(-d*x + 1))/sqrt(d*x + 1) - sqrt(d*x + 1)/(sq 
rt(2) - sqrt(-d*x + 1)))^5 - 3*b*d^3*((sqrt(2) - sqrt(-d*x + 1))/sqrt(d*x 
+ 1) - sqrt(d*x + 1)/(sqrt(2) - sqrt(-d*x + 1)))^5 - 16*a*d^4*((sqrt(2) - 
sqrt(-d*x + 1))/sqrt(d*x + 1) - sqrt(d*x + 1)/(sqrt(2) - sqrt(-d*x + 1)))^ 
3 + 6*c*d^2*((sqrt(2) - sqrt(-d*x + 1))/sqrt(d*x + 1) - sqrt(d*x + 1)/(sqr 
t(2) - sqrt(-d*x + 1)))^5 + 96*a*d^4*((sqrt(2) - sqrt(-d*x + 1))/sqrt(d*x 
+ 1) - sqrt(d*x + 1)/(sqrt(2) - sqrt(-d*x + 1))) - 48*c*d^2*((sqrt(2) - sq 
rt(-d*x + 1))/sqrt(d*x + 1) - sqrt(d*x + 1)/(sqrt(2) - sqrt(-d*x + 1)))^3 
+ 48*b*d^3*((sqrt(2) - sqrt(-d*x + 1))/sqrt(d*x + 1) - sqrt(d*x + 1)/(sqrt 
(2) - sqrt(-d*x + 1))) + 96*c*d^2*((sqrt(2) - sqrt(-d*x + 1))/sqrt(d*x + 1 
) - sqrt(d*x + 1)/(sqrt(2) - sqrt(-d*x + 1))))/(((sqrt(2) - sqrt(-d*x + 1) 
)/sqrt(d*x + 1) - sqrt(d*x + 1)/(sqrt(2) - sqrt(-d*x + 1)))^2 - 4)^3)/d
 

Mupad [B] (verification not implemented)

Time = 5.34 (sec) , antiderivative size = 304, normalized size of antiderivative = 3.07 \[ \int \frac {a+b x+c x^2}{x^4 \sqrt {1-d x} \sqrt {1+d x}} \, dx=\frac {b\,d^2\,\ln \left (\frac {{\left (\sqrt {1-d\,x}-1\right )}^2}{{\left (\sqrt {d\,x+1}-1\right )}^2}-1\right )}{2}-\frac {\frac {b\,d^2\,{\left (\sqrt {1-d\,x}-1\right )}^2}{{\left (\sqrt {d\,x+1}-1\right )}^2}-\frac {b\,d^2}{2}+\frac {15\,b\,d^2\,{\left (\sqrt {1-d\,x}-1\right )}^4}{2\,{\left (\sqrt {d\,x+1}-1\right )}^4}}{\frac {16\,{\left (\sqrt {1-d\,x}-1\right )}^2}{{\left (\sqrt {d\,x+1}-1\right )}^2}-\frac {32\,{\left (\sqrt {1-d\,x}-1\right )}^4}{{\left (\sqrt {d\,x+1}-1\right )}^4}+\frac {16\,{\left (\sqrt {1-d\,x}-1\right )}^6}{{\left (\sqrt {d\,x+1}-1\right )}^6}}-\frac {b\,d^2\,\ln \left (\frac {\sqrt {1-d\,x}-1}{\sqrt {d\,x+1}-1}\right )}{2}-\frac {c\,\sqrt {1-d\,x}\,\sqrt {d\,x+1}}{x}-\frac {\sqrt {1-d\,x}\,\left (\frac {2\,a\,d^3\,x^3}{3}+\frac {2\,a\,d^2\,x^2}{3}+\frac {a\,d\,x}{3}+\frac {a}{3}\right )}{x^3\,\sqrt {d\,x+1}}+\frac {b\,d^2\,{\left (\sqrt {1-d\,x}-1\right )}^2}{32\,{\left (\sqrt {d\,x+1}-1\right )}^2} \] Input:

int((a + b*x + c*x^2)/(x^4*(1 - d*x)^(1/2)*(d*x + 1)^(1/2)),x)
 

Output:

(b*d^2*log(((1 - d*x)^(1/2) - 1)^2/((d*x + 1)^(1/2) - 1)^2 - 1))/2 - ((b*d 
^2*((1 - d*x)^(1/2) - 1)^2)/((d*x + 1)^(1/2) - 1)^2 - (b*d^2)/2 + (15*b*d^ 
2*((1 - d*x)^(1/2) - 1)^4)/(2*((d*x + 1)^(1/2) - 1)^4))/((16*((1 - d*x)^(1 
/2) - 1)^2)/((d*x + 1)^(1/2) - 1)^2 - (32*((1 - d*x)^(1/2) - 1)^4)/((d*x + 
 1)^(1/2) - 1)^4 + (16*((1 - d*x)^(1/2) - 1)^6)/((d*x + 1)^(1/2) - 1)^6) - 
 (b*d^2*log(((1 - d*x)^(1/2) - 1)/((d*x + 1)^(1/2) - 1)))/2 - (c*(1 - d*x) 
^(1/2)*(d*x + 1)^(1/2))/x - ((1 - d*x)^(1/2)*(a/3 + (2*a*d^2*x^2)/3 + (2*a 
*d^3*x^3)/3 + (a*d*x)/3))/(x^3*(d*x + 1)^(1/2)) + (b*d^2*((1 - d*x)^(1/2) 
- 1)^2)/(32*((d*x + 1)^(1/2) - 1)^2)
 

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 204, normalized size of antiderivative = 2.06 \[ \int \frac {a+b x+c x^2}{x^4 \sqrt {1-d x} \sqrt {1+d x}} \, dx=\frac {-4 \sqrt {d x +1}\, \sqrt {-d x +1}\, a \,d^{2} x^{2}-2 \sqrt {d x +1}\, \sqrt {-d x +1}\, a -3 \sqrt {d x +1}\, \sqrt {-d x +1}\, b x -6 \sqrt {d x +1}\, \sqrt {-d x +1}\, c \,x^{2}-3 \,\mathrm {log}\left (-\sqrt {2}+\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-d x +1}}{\sqrt {2}}\right )}{2}\right )-1\right ) b \,d^{2} x^{3}+3 \,\mathrm {log}\left (-\sqrt {2}+\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-d x +1}}{\sqrt {2}}\right )}{2}\right )+1\right ) b \,d^{2} x^{3}-3 \,\mathrm {log}\left (\sqrt {2}+\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-d x +1}}{\sqrt {2}}\right )}{2}\right )-1\right ) b \,d^{2} x^{3}+3 \,\mathrm {log}\left (\sqrt {2}+\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-d x +1}}{\sqrt {2}}\right )}{2}\right )+1\right ) b \,d^{2} x^{3}}{6 x^{3}} \] Input:

int((c*x^2+b*x+a)/x^4/(-d*x+1)^(1/2)/(d*x+1)^(1/2),x)
 

Output:

( - 4*sqrt(d*x + 1)*sqrt( - d*x + 1)*a*d**2*x**2 - 2*sqrt(d*x + 1)*sqrt( - 
 d*x + 1)*a - 3*sqrt(d*x + 1)*sqrt( - d*x + 1)*b*x - 6*sqrt(d*x + 1)*sqrt( 
 - d*x + 1)*c*x**2 - 3*log( - sqrt(2) + tan(asin(sqrt( - d*x + 1)/sqrt(2)) 
/2) - 1)*b*d**2*x**3 + 3*log( - sqrt(2) + tan(asin(sqrt( - d*x + 1)/sqrt(2 
))/2) + 1)*b*d**2*x**3 - 3*log(sqrt(2) + tan(asin(sqrt( - d*x + 1)/sqrt(2) 
)/2) - 1)*b*d**2*x**3 + 3*log(sqrt(2) + tan(asin(sqrt( - d*x + 1)/sqrt(2)) 
/2) + 1)*b*d**2*x**3)/(6*x**3)